First find the slope of the reference line.
slope=m=(y2-y1)/(x2-x1) for any two points (x,y)
In this case we have points (-3,2) and (0,1) clearly marked.
m=(1-2)/(0--3)
m=-1/3
For two lines to be perpendicular their slopes must be negative reciprocals of one another, mathematically:
m1*m2=-1, in this case our reference line has a slope of -1/3 so our perpendicular line must have a slope that satisifies:
-m/3=-1
m/3=1
m=3, so in the slope-intercept form of a line, y=mx+b, we so far have:
y=3x+b, using any point we can now solve for b, the y-intercept (the value of y when x=0), We are told that we must pass through the point (3,4) so:
4=3(3)+b
4=9+b
-5=b, now we know the complete equation of the perpendicular line passing through (3,4) is:
y=3x-5